The area of the triangle formed by points of intersection of parabola y=a(x−3)(x+2) with the coordinate axes is 10. Find a if it is known that parabola opens upward.

a = 4

Step-by-step explanation: Area of a triangle is calculated as: .

The triangle formed by the parabola has base (b) equal to the distance between the points where the graph touches x-axis and height (h) is the point where graph touches the y-axis.

The points on the x-axis are the roots of the quadratic equation:

a(x-3)(x+2)=0

(x-3)(x+2)=0

x - 3 = 0

x = 3

or

x + 2 = 0

x = -2

So, base is the distance between (-2,0) and (3,0).

Since they are in the same coordinate, distance will be:

b = 3 - (-2)

b = 5

Area of the triangle is 10. So constant a is

5a = 10.2

a = 4

The constant a of the function y = a(x-3)(x+2) is 4.

f(x) is the value of the function. b is the value of the function when x equals zero or the y-coordinate of the point where the line crosses the y-axis in the coordinate plane.

step-by-step explanation:

hope this !

what i would say the best answer is a. 40.5